The Bernoulli Equation

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1 The Bernoulli Equation The most used and the most abused equation in fluid mechanics. Newton s Second Law: F = ma In general, most real flows are 3-D, unsteady (x, y, z, t; r,θ, z, t; etc) Let consider a -D motion of flow along streamlines, as shown below. Velocity (V ): Time rate of change of the position of the particle. Streamlines: The lines that are tangent to the velocity vectors throughout the flow field. Note: For steady flows, each particle slide along its path, and its velocity vector is everywhere tangent to the path. Streamline coordinate: v S = S(t) ; V = ds / dt (the distance along the streamline can be decided by V and R(s) )

By chain rule, a s = dv = V ds = V V ; a dt S dt S where R is the local radius of curvature of the streamline, and S is the distance measured along the streamline from some arbitrary initial point.

2 By chain rule, a s = dv = V ds = V V ; a dt S dt S where R is the local radius of curvature of the streamline, and S is the distance measured along the streamline from some arbitrary initial point. n = V R Figure 3. Isolation of small fluid particle in a flow field. V 0; S R (not straight line) Forces: Gravity & Pressure (important) v F = ma v Viscous, Surface tension (negligible) along a streamline

Figure 3.3 Free-body diagram of a fluid particle for which the important forces are those due to pressure and gravity.. Consider the small fluid particle (δ S δ n ), as show above.

3 Figure 3.3 Free-body diagram of a fluid particle for which the important forces are those due to pressure and gravity.. Consider the small fluid particle (δ S δ n ), as show above. If the flow is steady. N. δ F S = δ m a S = δ mv V S = ρ δ V V S (3.) (Eq. (3.) is valid for both compressible and incompressible fluids) W. δ W S = δ W sin θ = γ δ sin θ (θ = 0 δ W S = 0)

4 δ P P. S P δ S S (first term of Taylor series expansion, because the particle is small; P + δ P S P δ P S ) Thus δ F P S δ F PS : the net pressure force on the particle in the streamline direction = (P δ P S )δ nδ y (P + δ P S )δ nδ y = δ P δ nδ y = P δ sδ nδ y S S = P δ S *Note: if pressure gradient is not zero ( P C ), then there is a net pressure force. P = P S v + S P n v n Net Force δ F S = δ W S + δ F PS ; ρ V V S = γ sin θ P S (3.4) Note: Force balancing consideration ρ V (mass flux), not ρ or V, is a key parameter for fluid mechanics, if ρ constant.

5 Example 3.1 Consider the inviscid, incompressible, steady flow along the horizontal streamline A-B in front of the sphere of radius a as shown in Fig. E3.1a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is

6 3 V = V (1 + a ) 0 x 3 Determinate the pressure variation along the streamline from point A far in front the sphere (x A = and V A = V 0 ) to point B on the sphere ( x B = a and V B = 0)

7 Figure E3.1 Solution: Since the flow is steady and inviscid, Eq. 3.4 is valid. In addition, since the streamline is horizontal, sin θ = sin 0 = 0 and the equation of motion along the streamline reduces to p = ρ V V s s (1) with the given velocity variation along the streamline, the acceleration term is V V s = V V x = V 0 (1 + 3 a 3 x 3 a a = 3V 0 (1 + ) x 3 x 4 3 )( 3V 0a 3 ) x

8 where we have replaced s by x since the two coordinates are

9 identical (within an additive constant) along streamline A-B. It follows that V V s < along the streamline. The fluid slows 0 down from V 0 far ahead of the sphere to zero velocity on the nose of the sphere (x = a). Thus according to Eq. 1, to produce the given motion the pressure gradient along the streamline is p 3 a 3 3ρ a V 0 (1 + x 3 ) x 4 () = x This variation is indicated in Fig. E3. 1b. It is seen that the pressure increase in the direction of flow ( p x > from point A to point B. The maximum pressure gradient 0) (0.610 ρ V a) 0 occurs just slightly ahead of the sphere (x = 1.05a). It is the pressure gradient that slows the fluid down from V A = V 0 to V B = 0. The pressure distribution along the streamline can be obtained by integrating Eq. from p = 0 (gage) at x = to pressure p at location x. The result, plotted in Fig. E3.1c, is a (a x) 6 3

10 p = ρ V ( ) + 0 x (Ans) The pressure at B, a stagnation point since V B = 0, is the

11 highest pressure along the streamline ( p B = ρ V 0 ). As shown in Chapter 9, this excess pressure on the front of the sphere (i.e. p B > 0 ) contributes to the net drag force on the sphere. Note that the pressure gradient and pressure are directly proportional to the density of the fluid, a representation of the fact that the fluid inertia is proportional to its mass. Since sin θ = dz dv 1 d(v ) ds, also V = ds ds streamline, n = const, so dn = 0., and along the Thus dp = p ds + s p dn = n dp ds ds Eq. (3.4) γ dz dp = 1 ρ d(v ) ds ds ds along a streamline (s) dp + 1 ρ d(v ) + γ dz = 0 dp 1 or + V + gz = const ρ (3.5) (3.6) Note: if ρ constant, then ρ = f ( p)

12 must be known. The well-known Bernoulli equation:

13 P + 1 ρ V + γ Z = constant Note: 4 assumptions 1) μ = 0 (inviscid) ) = 0 t 3) ρ = C (steady) (incompressible) 4) along a streamline (dn = 0 ) Example 3. Consider the flow of air around a bicyclist moving through still air with velocity V 0 as is shown in Fig. E3.. Determine the difference in the pressure between points (1) and (). Figure E 3. Solution: In a coordinate system fixed to the bike, it appears as through the air is flowing steadily toward the bicycle with speed

14 V 0. If the assumptions of Bernoulli s equation are valid (steady,

15 incompressible, inviscid flow), Eq. 3.7 can be applied as follows along the streamline that passes through (1) and () p ρ V + γ z 1 p + 1 ρ V + γ z 1 = We consider (1) to be in the free stream so that V 1 = V 0 and () to be at the tip of the bicyclist s nose and assume that z 1 = z and V = 0 (both of which, as is discussed in Section 3.4, are reasonable assumptions). It follows that the pressure at () is greater than at (1) by an amount p p 1 = 1 ρ V = 1 ρ V 1 0 (Ans) A similar result was obtained in Example 3.1 by integrating pressure gradient, which was known because the velocity distribution along the streamline, V (s), was known. The Bernoulli equation is a general integration of F = ma. To determine p p 1, knowledge of the detailed velocity distribution is not needed-only the boundary conditions at (1) and () are required. Of course, knowledge of the value of V along the streamline is needed to determine the speed V 0. As discussed in Section 3.5, this is the principle upon which many

16 velocity measuring devices are based. If the bicyclist were accelerating or decelerating, the flow would unsteady (i.e. V 0 constant) and the above analysis would be in correct since Eq. 3.7 is restricted to steady flow. v v F = ma normal to a streamline. Example: The devastating low-pressure region at the center of a tornado can be explained by applying Newton s nd law across the nearly circular streamlines of the tornado.. δ Fn = δ mv R. δ Wn = δ W cosθ = ρ δ V R = δ m a v = γ δ cos θ n (3.8) (θ = 90 ;δ W n = 0). δ Fpn = (P δ Pn)δ Sδ y (P + δ Pn)δ Sδ y = δ Pnδ Sδ y = P δ Sδ nδ y n. Thus, = P δ n δ Fn = δ Wn + δ Fpn = cos θ P (3.9) ( γ )δ n Eq. (3.8) & (3.9) &cosθ = dz dn

17 γ dz P = dn n ρ V R (3.10)

18 . if gravity is neglected or if the floor is horizontal P n ρ V γ = 0 dz = 0 dn = R This pressure difference is needed to be balance the centrifugal acceleration associated with the curved streamlines of the fluid motion.. Read example 3.3. Integrate across the streamline, using the factor that P n = dp dn if S = constant dp V ρ + dn + gz = constant (3.11) R across the streamline v V = V (S,n) ; R = R(S,n) p + ρ V dn + γ z = C R if 1) t ) μ = 0 3) ρ = C (3.1)

19 Example 3.3 Shown in Figs. E3.3a, b are two fields with circular streamlines. The velocity distributions are V (r) = C 1 r V (r) = C r for case (a) for case(b) and where C 1 and C are constant. Determine the pressure distributions, p = p(r), for each given that p = p 0 at r = r 0.

20 Figure E 3.3 Solution: We assume the flow are steady, inviscid, and incompressible with streamline in the horizontal plane ( dz dn = 0) Since the streamlines are circles, the coordinate n points in a direction opposite of that of the radial coordinate, n = r, and the radius of curvature is given by R = r. Hence, Eq becomes p ρ V = r For case (a) this gives r p = ρ C 1 r r p ρ C While for case (b) it gives = r r 3 For either the pressure increases as r increase since δ p δ r > 0 Integration of these equations with respect to r, starting with a known pressure p = p 0 at r = r 0, gives

21 p = 1 ρ C (r r 1 0 ) + p 0 for case (a) and (Ans) p = 1 ρ C ( 1 1 ) + p r r 0 0 (Ans) for case (b). These pressure distributions are sketched in Fig E3.3c. The pressure distributions needed to balance the centrifugal accelerations in cases (a) and (b) are not the same because the velocity distributions are different. In fact for case (a) the pressure increase without bound as r, while for case (b) the pressure approaches a finite value as r. The streamline patterns are the same for each case, however. Physically, case (a) represents rigid body rotation (as obtained in a can of water on a turntable after it has been spun up ) and case(b) represents a free vortex (an approximation to a tornado or the swirl of water in a drain, the bathtub vortex ).

22 Physical Interpretation. Along a streamline: P + 1 ρ V + γ z = C Across the streamline: P + ρ V R dn + γ z = c If: steady, inviscid, and incompressible (None is exactly true for real flows). Physics: Force balance; Bernoulli s Principle The work done on a particle by all forces acting on the particle is equal to the change of the kinetic energy of the particle. (Newton s second law; first & second laws of thermodynamics). Example 3.4 Consider the flow of water from the syringe shown in Fig A force applied to the plunger will produce a pressure greater than atmospheric at point (1) within the syringe. The water flows from the needle, point (), with relatively high velocity and coasts up to point (3) at the top of its trajectory. Discuss the

23 energy of the fluid at points (1), (), and (3) by using the Bernoulli equation. Figure E 3.4 Solution: If the assumptions (steady, inviscid, incompressible flow) of the Bernoulli equation are approximately valid, it then follows that the flow can be explained in terms of the partition of the total energy of the water. According to Eq the sum of the three types of energy (kinetic, potential, and pressure) or heads (veloci ty, elevation, and pressure) must remain constant. The following table indicates the relative magnitude of each of these energies at the three points shown in the figure.

24 Energy Type Point Kinetic ρ V Potential γ z Pressure p 1 Small Zero Large Large Small Zero 3 Zero Large Zero The motion results in (or is due to) a change in the magnitude of each type of energy as the fluid flows from one location to another. An alternate way to consider this flow is as follows. The pressure gradient between (1) and () produces an acceleration to eject the water from the needle. Gravity acting on the particle between () and (3) produces a deceleration to cause the water to come to a momentary stop at the top of its flight. If friction (viscous) effects were important, there would be an energy loss between (1) and (3) and for the given p 1 the water would not be able to reach the height indicated in the figure. Such friction may arise in the needle (see chapter 8, pipe flow) or between the water stream and the surrounding air (see chapter

9, external flow). Example 3.5 Consider the inviscid, incompressible, steady flow shown in Fig. E3.5. From section A to B the streamlines are straight, while from C to D they follow circular paths.

25 9, external flow). Example 3.5 Consider the inviscid, incompressible, steady flow shown in Fig. E3.5. From section A to B the streamlines are straight, while from C to D they follow circular paths. Describe the pressure variation between points (1) and () and points (3) and (4). Figure E 3.5 Solution: With the above assumption and the fact that R = for the portion from A to B, Eq becomes p + γ z = constant The constant can be determined by evaluating the known variables at the two locations using P = 0 (gage), Z 1 = h -1 to give

26 p 1 = p + γ ( z z 1 ) = p + γ h 1 (Ans) Note that since the radius of curvature of the streamline is infinite, the pressure variation in the vertical direction is the same as if the fluid were stationary. However, if we apply Eq between points (3) band point (4) we obtain (using dn = -dz) V p 4 + ρ Z 4 ( dz ) + Z 3 R γ z 4 = p 3 + γ z 3 With p 4 =0 and z 4 -z 3 =h 4-3 this becomes p h = γ Z dz ρ Z 4 V 3 R (Ans) To evaluate the integral we must know the variation of V and R with z. Even without this detailed information we note that the integral has a positive value. Thus, the pressure at (3) is less than the hydrostatic value, γ h 4, by an amount equal to 3 ρ dz Z V 4 Z 3 R. This lower pressure, caused by the curved streamline, is necessary to accelerate the fluid around the curved path.

27 Note that we did not apply the Bernoulli equation (Eq. 3.13)

28 across the streamlines from (1) to () or (3) to (4). Rather we used Eq As is discussed in section 3.6 application of the Bernoulli equation across streamlines (rather than along) them may lead to serious errors. 3.5 Static, Stagnation (Total), and Dynamic Pressure γ z - hydrostatic pressure not a real pressure, but possible due to potential energy variations of the fluid as a result of elevation changes. p - static pressure 1 ρ V -dynamic pressure stagnation point V = 0 P = P + 1 ρ V stagnation pressure static pressure+dynamic pressure (if elevation effect are neglected) Total pressure It represents the conversion of all of the kinetic energy into a pressure rise.

P T = γ Z + 1 + ρ V P Total P Hydrostatic P Static P Dynamic P Figure 3.

29 P T = γ Z ρ V P Total P Hydrostatic P Static P Dynamic P Figure 3.4 Measurement of static and stagnation pressures Pitot-static tube: simple, relatively in expensive.

30 Figure 3.6 The Pitot-static tube Figure 3.7 Typical Piotot-static tube designs. p + 1 ρ V + γ z = p T = constant along a streamline p = p + 1 ρ V center tube: 3 elevation is neglected. outer tube: p 4 = p 1 = p p p = 1 ρ V V 3 4 = ( p p ) / ρ 3 4

31 Figure 3.9 Typical pressure distribution along a Pitot-static tube. The cylinder is rotated until the pressures in the two sideholes are equal, thus indicating that the center hole points directly upstream-measure stagnation pressure. Side holes ( β = 9.5 ) measure static pressure. Figure 3.10 Cross section of a directional-finding Pitot-static tube.

32 3.6 Example of Use of the Bernoulli Equation steady ( t = 0), inviscid ( μ = 0 ), incompressible ( ρ = constant ) and along a streamline ( n = 0). Between two points: P + 1 ρ V 1 + γ Z 1 1 = P + 1 ρ V + γ Z Unknowns: P 1, P, V 1, V, Z 1, Z (6) If 5 unknowns are given, then determining the remaining one. Free Jets (Vertical flows from a tank) (1) - ():

33 Figure 3.11 Vertical flow from a tank (1) - ():

34 P 1 = 0 (gage pressure); P = 0 (Free jet) Z 1 = h; Z = 0; V 1 0; γ h = 1 ρ V V = γ h = ρ gh () and (4): atmospheric pressure (5): V 5 = g(h + H ) (3) and (4): P 3 = γ (h l) ; = 0 ; Z 3 P 4 1 V 3 =0 P 3 + γ l = P 4 + ρ V 4 = l ; Z 4 = 0, Q V 4 = gh Recall from Physics or dynamics 自由落體在一真空中 V = gh ; the same as the liquid leaving from the nozzle. All potential energy kinetic energy (neglecting the viscous effects).

35 Fig 3.1 Horizontal flow from a tank. Vena contracta effect: Fluid can not turn 90 ; Contraction coefficient: C = A /A ; c j h A j cross area of jet fluid column; A h cross area of the nozzle exit.

36 Figure 3.14 Typical flow patterns and contraction coefficients for various round exit configurations. Confined flows Figure 3.15 Steady flow into and out of a tank.

37 mass flow rate m = ρ Q (kg/s or slugs/s) where Q: volume flow rate (m 3 /s or ft 3 /s) Q = VA ; m = ρ Q = ρ VA From Fig. 3.15: ρ 1 A 1 V 1 = ρ A V Example 3.7 A stream of diameter d = 0.1m flows steadily from a tank of diameter D = 1.0m as shown in Fig E3.7a. Determine the flow rate, Q, needed from the inflow pipe if the water depth remains constant. h =.0m.

38 Figure E 3.7 Solution: For steady, inviscid, incompressible the Bernoulli equation applied between points (1) and () is p + 1 ρ V 1 + γ z = p ρ V + γ Z (1) With the assumptions that p 1 = p = 0, z 1 = h, and z = 0, Eq. 1 becomes 1 V + gh = 1 V () 1 Although the water level remains constant (h = constant), there is an average velocity, V 1, across section (1) because of the flow

39 from the tank. From Eq 3.19 for steady incompressible flow, conservation of mass requires Q 1 = Q = AV. Thus, A 1 V 1 = A V, or Hence, π D V = π d V d V 1 = ( ) V (3) D Equation 1 and 3 can be combined to give V = gh 1 (d / D) 4 Thus, with the given data V = (9.81m/s )(.0 m) 1 (0.1m/1m) 4 = 6.6 m/s and Q = AV = A V = π (0.1m) (6.6m/s) = 0.049m 3 /s (Ans) In this example we have not neglected the kinetic energy in the water in the tank (V 1 0). If the tank diameter is large compared to the jet diameter (D>>>d), Eq. 3 indicates that V 1 <<V and the assumption that V 1 0 would be reasonable. The error associated with this assumption can be seen by calculating the ratio of the flowrate assuming V 1 0, denoted Q, to that assuming V 1 = 0, denoted Q 0. This ratio, written as

Q V gh (1 (d D) 4 ) 1 = = = Q 0 V D= gh 1 (d D) 4 is plotted in Fig. E3.7b. With 0< d/d < 0.4 it follows that 1< Q/Q 0 1.01, and the error in assuming V 1 = 0 is less than 1%.

40 Q V gh (1 (d D) 4 ) 1 = = = Q 0 V D= gh 1 (d D) 4 is plotted in Fig. E3.7b. With 0< d/d < 0.4 it follows that 1< Q/Q , and the error in assuming V 1 = 0 is less than 1%. Thus, it is often reasonable to assume V 1 = 0. Example 3.9 Water flows through a pipe reducer as is shown in Fig. E3.9. The static pressure at (1) and () are measured by the inverted U-tube manometer containing oil of specific gravity. SG. Less than one. Determine the manometer reading, h. FIGURE E 3.9 Solution: With the assumption of steady, inviscid, incompressible flow, the Bernoulli equation can be written as

41 p + 1 ρ V +γ z = p + 1 ρ V +γ z The continuity equation (Eq. 3.19) provides a second relationship between V 1 and V, if we assume the velocity profiles are uniform as those two locations and the fluid incompressible: Q = AV = AV 1 1 By combining these two equations we obtain p p = γ ( z ] z ) + 1 ρ V [1 A (1) 1 1 A1 This pressure difference is measured by the manometer and can be determined by using the pressure-depth ideas developed in Chapter. Thus. p γ ( z z ) l γ h + SGγ h +γ l = p 1 1 γ or p p 1 = γ ( z z 1 ) + (1 SG )γ h () As discussed in Chapter, the pressure difference is neither merely γh not γ(h+z 1 -z ). Equations 1 and can be combined to give the desired result as follows (1 SG )γ h = 1 ρ V [1 A ] A1

42 A 1 Q A or V =Q/A, thus, h = 1. (Ans) A g(1 SG ) The difference in elevation, z 1 -z, was not needed because the change in elevation term in the Bernoulli equation exactly cancels the elevation term in the manometer equation. However, the pressure difference, p 1 -p, depends on the angle θ, because of the elevation, z 1 -z, in Eq. 1. Thus, for a given flowrate, the pressure difference, p 1 -p, as measured by a pressure gage would vary with θ, but the manometer reading, h, would be independent of θ. In general, an increase in velocity is accompanied by a decrease in pressure. Air (gases): Compressibility (Ch 11) Liquids: Cavitation (Propeller etc.) Flow Rate Measurement z Ideal flow meter neglecting viscous and compressible effects. (loss some accuracy)

43 z Orifice meter; Nozzle meter: V, P ; Venturi meter Q = ( P P ) A ρ 1 A 1 1 A (3.0) 1 z Sluice gate: Q = z b g( z z ) z 1 z 1 (3.1) z Sharp crested weir: Q = C Hb gh = C b 1 1 gh 3 The Energy Line and the Hydraulic Grade Line z Bernoulli equation is an energy equation with four assumptions: inviscid, incompressible, steady flow, and along a streamline from pts ( 1) - (), respectively, and the sum of partition of energy remains constant from pts (1)-(). (Head) H = p + V + z = constant along a streamline. γ g

grade line The energy line and hydraulic grade line

44 FIGURE 3.1 Representation of the energy line and the hydraulic grade line FIGURE 3. The energy line and hydraulic grade line for flow from a tank. Restrictions on the use of Bernoulli Equation Compressibility Effect

45 Gases: dp ρ not a constant A special case: for compressible flow Given - steady, inviscid, isothermal (T = C along the streamline) Sol: p = ρ RT ρ dp = RT ρ = p RT dp = RT ln p + c p V V RT p 1 + = + z ln 1 z + g g g p1 V 1 or + z + 1 g RT p V ln 1 = + z g p g Compare to the incompressible Bernoulli Equation. Q if p p p 1 = = 1 +ε, ε = p p 1 << 1 p p p p then ln 1 = ln(1 + ε ) ε p V RT p 1 + z + ( 1 1) = + z 1 g g p g V p V p + + V 1 z 1 + g 1 γ = + g z γ p p p = 1 = 1 1 p p 3.8. Unsteady Effect

46 V = V(s) steady; V = V(s, t) unsteady.

47 a s t = V +V V s (local + convective accelerations) z Repeating the steps leading to the Bernoulli Equation. ρ V ds + dp + 1 ρ d (V ) +γ dz = 0 t along a streamline z If incompressible, inviscid, p S v 1 ρ V +γ z = ρ 1 1 S ds + + ρ V +γ p 1 1 t z Use velocity potential to simplify the problem (Ch 6). FIGURE 3.5 Oscillation of a liquid column in a U-tube

48 3.8.3 Rotational Effect In general, the Bernoulli constant varies from streamline to streamline. However, under certain restrictions, this constant may be the same throughout the entire flow field, as ex.3.19 illustrates this effect Other restrictions μ = 0 (inviscid); Bernoulli Equation: A first integral of Newton s nd laws along a streamline.

FLUID MECHANICS. Chapter 3 Elementary Fluid Dynamics - The Bernoulli Equation

FLUID MECHANICS. Chapter 3 Elementary Fluid Dynamics - The Bernoulli Equation FLUID MECHANICS Chapter 3 Elementary Fluid Dynamics - The Bernoulli Equation CHAP 3. ELEMENTARY FLUID DYNAMICS - THE BERNOULLI EQUATION CONTENTS 3. Newton s Second Law 3. F = ma along a Streamline 3.3